If `int(dx)/(sqrt(sin^3xcos^5x))=asqrt(cotx)+bsqrt(tan^3)x+c ,` then `a=-1,b=1/3` (b) `a=-3,b=2/3` `a=-2,b=4/3` (d) none of these

Options

A. `a= -1,b=1//3`
B. `a= -3,b=2//3`
C. `a= -2,b=4//3`
D. `a= -2, b=2//3`

Answer

D

Explanation

`I=int(dx)/(sqrt(sin^(3)xcos^(5)x))`
`=int(dx)/(sqrt((sin^(3)x)/(cos^(3)x)cos^(8)x))`
`=int(sec^(4)x)/(sqrt(tan^(3)x))dx`
`=int((1+tan^(2)x)sec^(2)x)/(sqrt(tan^(3)x))dx`
Let `t=tanx " or "dt=sec^(2)x dx`
` :. I=int(1+t^(2))/(t^(3//2))dt`
`=int(t^(-3//2)+t^(1//2))dt`
`= -2t^(-(1)/(2))+(2)/(3)t^(3//2)+C`
`= -2 sqrt(cotx)+(2)/(3)sqrt(tan^(3)x)+C`
` :. a= -2,b=(2)/(3)`